This error is usually because a variable is being used in a preg_* function. Usually placing the variable (denated with a $ in front of a word like $variable) inside of preg_quote() fixes this. For example $variable would become preg_quote($variable). Without seeing the code on that line though its hard to say for sure.

EDIT: I should also note that improper information being fed to this variable could cause this as well and it may not be an actual bug per say.